Woman who weighs 500 N stands on an 8.0-m-long board that weighs 100 N. The board is supported at each end. The support force at the right end is 3 times the support force at the left end. How far from the right end is the woman standing

Explanation:

Let us assume that W is the weight of the board and is the weight of women standing.

And, at the left end supporting force is F and at the right end supporting force is 3F.

So, when this system is in equilibrium then upward force will be equal to the downward force as follows.

              4F = 100 N + 500 N

                 F =

                    = 150 N

For system to be in static friction another condition is that net torque acting on the left end will be equal to zero.

Hence,         = 0

          = 0

Putting the given values into the above formula as follows.

          = 0

        = 0  

                    x = 6.4 m

Therefore, at the left side distance from the women is 6.4 m.

And, distance from the right end will be as follows.

               8 m - 6.4 m = 1.4 m

that would be because of both gravity and friction. so     c

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards